Real Analysis: Part I. William G. Faris

Transcription

1 Real Analysis: Part I William G. Faris February 2, 2004

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3 Contents 1 Mathematical proof Logical language Free and bound variables Proofs from analysis Natural deduction Natural deduction strategies Equality Lemmas and theorems More proofs from analysis Sets Zermelo axioms Comments on the axioms Ordered pairs and Cartesian product Relations and functions Number systems Relations, Functions, Dynamical Systems Identity, composition, inverse, intersection Picturing relations Equivalence relations Generating relations Ordered sets Functions Relations inverse to functions Dynamical systems Picturing dynamical systems Structure of dynamical systems Functions, Cardinal Number Functions Picturing functions Indexed sums and products Cartesian powers iii

4 iv CONTENTS 4.5 Cardinality Ordered sets and completeness Ordered sets Order completeness Sequences in a complete lattice Order completion The Knaster-Tarski fixed point theorem The extended real number system Metric spaces Metric space notions Normed vector spaces Spaces of finite sequences Spaces of infinite sequences Spaces of bounded continuous functions Open and closed sets Continuity Uniformly equivalent metrics Sequences Metric spaces and completeness Completeness Uniform equivalence of metric spaces Completion The Banach fixed point theorem Coerciveness Metric spaces and compactness Total boundedness Compactness Countable product spaces Compactness and continuous functions Semicontinuity Compact sets of continuous functions Curves of minimum length Vector lattices Positivity Integration of regulated functions The Riemann integral Step functions Coin tossing Vector lattices Elementary integrals Integration on a product of finite spaces

5 CONTENTS v 10 The integral The Daniell construction Stage one Stage two Example: Coin tossing Example: Lebesgue measure Measurable functions Monotone classes Generating monotone classes Sigma-algebras of functions Generating sigma-algebras Sigma-rings of functions Rings and algebras of sets The integral on measurable functions Integration Uniqueness of integrals Existence of integrals Probability and expectation Image integrals The Lebesgue integral The Lebesgue-Stieltjes integral Integrals on a σ-ring Integrals and measures Terminology Convergence theorems Measure Extended real valued measurable functions Fubini s theorem for sums and integrals Fubini s theorem for sums Fubini s theorem Introduction Sigma-finite integrals Summation Product sigma-algebras The product integral Tonelli s theorem Fubini s theorem Semirings and rings of sets

6 vi CONTENTS 15 Probability Coin-tossing Weak law of large numbers Strong law of large numbers Random walk

7 Chapter 1 Mathematical proof 1.1 Logical language There are many useful ways to present mathematics; sometimes a picture or a physical analogy produces more understanding than a complicated equation. However, the language of mathematical logic has a unique advantage: it gives a standard form for presenting mathematical truth. If there is doubt about whether a mathematical formulation is clear or precise, this doubt can be resolved by converting to this format. The value of a mathematical discovery is considerably enhanced if it is presented in a way that makes it clear that the result and its proof could be stated in such a rigorous framework. Here is a somewhat simplified model of the language of mathematical logic. There may be function symbols. These may be 0-place function symbols, or constants. These stand for objects in some set. Example: 8. Or they may be 1-place functions symbols. These express functions from some set to itself, that is, with one input and one output. Example: square. Or they may be 2-place function symbols. These express functions with two inputs and one output. Example: +. Once the function symbols have been specified, then one can form terms. The language also has a collection of variables x, y, z, x, y, z,.... Each variable is a term. Each constant c is a term. If t is a term, and f is a 1-place function symbol, then f(t) is a term. If s and t are terms, and g is a 2-place function symbol, then g(s, t) or (sgt) is a term. Example: In an language with constant terms 1, 2, 3 and 2-place function symbol + the expression (x + 2) is a term, and the expression (3+(x+2)) is a term. Note: Sometimes it is a convenient abbreviation to omit outer parentheses. Thus 3 + (x + 2) would be an abbreviation for (3 + (x + 2)). The second ingredient is predicate symbols. These may be 0-place predicate symbols, or propositional symbols. They may stand for complete sentences. One useful symbol of this nature is, which is interpreted as always false. Or they may be 1-place predicate symbols. These express properties. Example: even. 1

8 2 CHAPTER 1. MATHEMATICAL PROOF Or they may be 2-place predicate symbols. These express relations. Example: <. Once the terms have been specified, then the atomic formulas are specified. A propositional symbol is an atomic formula. If p is a property symbol, and t is a term, then tp is an atomic formula. If s and t are terms, and r is a relation symbol, then srt is an atomic formula. Thus (x + 2) < 3 is an atomic formula. Note: This could be abbreviated x + 2 < 3. Finally there are logical symbols,,,,, and parentheses. Once the atomic formulas are specified, then the other formulas are obtained by logical operations. If A and B are formulas, then so are (A B), (A B), and (A B). If x is a variable and A(x) is a formula, then so are x A(x) and x A(x). Thus x x + 2 < 3 is a formula. We shall often abbreviate (A ) by A. Thus facts about negation will be special cases of facts about implication. In writing a formula, we often omit the outermost parentheses. However this is just an abbreviation. Another useful abbreviation is (A B) for ((A B) (B A)). Some of the logical operations deserve special comment. The implication A B is also written if A, then B A only if B B if A. The equivalence A B is also written A if and only if B. The converse of A B is B A. The contrapositive of A B is B A. When A is defined by B, the definition is usually written in the form A if B. It has the logical force of A B. The universal quantified formula x A(x) is also written for all x A(x) for each x A(x) for every x A(x). The existential quantified formula x A(x) is also written there exists x with A(x) for some x A(x). Note: Avoid at all cost expressions of the form for any x A(x). The word any does not function as a quantifier in the usual way. For example, if one says z is special if and only if for any singular x it is the case that x is tied to z, it is not clear which quantifier on x might be intended. Often a quantifier has a restriction. The restricted universal quantifier is x (C(x) A(x)). The restricted existential quantifier is x (C(x) A(x)). Here C(x) is a formula that places a restriction on the x for which the assertion is made. It is common to have implicit restrictions. For example, say that the context of a discussion is real numbers x. There may be an implicit restriction x R. Since the entire discussion is about real numbers, it may not be necessary to

9 1.2. FREE AND BOUND VARIABLES 3 make this explicit in each formula. This, instead of x (x R x 2 0) one would write just x x 2 0. Sometimes restrictions are indicated by use of special letters for the variables. Thus often i, j, k, l, m, n are used for integers. Instead of saying that m is odd if and only if y (y N m = 2y + 1) one would just write that m is odd if and only if k m = 2k + 1. The letters ɛ, δ are used for strictly positive real numbers. The corresponding restrictions are ɛ > 0 and δ > 0. Thus instead of writing x (x > 0 y (y > 0 y < x)) one would write ɛ δ δ < ɛ;. Other common restrictions are to use f, g, h for functions or to indicate sets by capital letters. Reasoning with restricted variables should work smoothly, provided that one keeps the restriction in mind at the appropriate stages of the argument. 1.2 Free and bound variables In a formula each occurrence of a variable is either free or bound. The occurrence of a variable x is bound if it is in a subformula of the form x B(x) or x B(x). (There may also be other operations, such as the set builder operation, that produce bound variables.) If the occurrence is not bound, then it is said to be free. In general, a bound variable may be replaced by a new bound variable without changing the meaning of the formula. Thus, for instance, if y is a variable that does not occur in the formula, one could replace the occurrences of y in the subformula y B(y) by y, so the new subformula would now be y B(y ). Of course if the variables are restricted, then the change of variable should respect the restriction. Example: Let the formula be y x < y. This says that there is a number greater than x. In this formula x is free and y is bound. The formula y x < y has the same meaning. In this formula x is free and y is bound. On the other hand, the formula y x < y has a different meaning. This formula says that there is a number greater than x. We wish to define careful substitution of a term t for the free occurrences of a variable x in A(x). The resulting formula will be denoted A(t) There is no particular problem in defining substitution in the case when the term t has no variables that already occur in A(x). The care is needed when there is a subformula in which y is a bound variable and when the term t contains the variable y. Then mere substitution might produce an unwanted situation in which the y in the term t becomes a bound variable. So one first makes a change of bound variable in the subformula. Now the subformula contains a bound variable y that cannot be confused with y. Then one substitutes t for the free occurrences of x in the modified formula. Then y will be a free variable after the substitution, as desired. Example: Let the formula be y x < y. Say that one wished to substitute y+1 for the free occurrences of x. This should say that there is a number greater

10 4 CHAPTER 1. MATHEMATICAL PROOF than y + 1. It would be wrong to make the careless substitution y y + 1 < y. This statement is not only false, but worse, it does not have the intended meaning. The careful substitution proceeds by first changing the original formula to y x < y. The careful substitution then produces y y + 1 < y. This says that there is a number greater than y + 1, as desired. The general rule is that if y is a variable with bound occurrences in the formula, and one wants to substitute a term t containing y for the free occurrences of x in the formula, then one should change the bound occurrences of y to bound occurrences of a new variable y before the substitution. This gives the kind of careful substitution that preserves the intended meaning. 1.3 Proofs from analysis The law of double negation states that A A. De Morgan s laws for connectives state that (A B) ( A B) and that (A B) ( A B). De Morgan s laws for quantifiers state that x A(x) x A(x) and x A(x) x A(x). Since (A B) (A B) and (A B) (A B), De Morgan s laws continue to work with restricted quantifiers. Examples: 1. The function f is continuous if a ɛ δ x( x a < δ f(x) f(a) < ɛ). It is assumed that a, x, ɛ, δ are real numbers with ɛ > 0, δ > The function f is not continuous if a ɛ δ x( x a < δ f(x) f(a) < ɛ). This is a mechanical application of De Morgan s laws. Similarly, the function f is uniformly continuous if ɛ δ a x( x a < δ f(x) f(a) < ɛ). Notice that the only difference is the order of the quantifiers. Examples: 1. Consider the proof that f(x) = x 2 is continuous. The heart of the proof is to prove the existence of δ. The key computation is x 2 a 2 = x + a x a = x a + 2a x a. If x a < 1 then this is bounded by (2 a + 1) x a. Here is the proof. Let ɛ > 0. Suppose x a < min(1, ɛ/(2 a + 1)). From the above computation it is easy to see that x 2 a 2 < ɛ. Hence x a < min(1, ɛ/(2 a + 1)) x 2 a 2 < ɛ. Since in this last statement x is arbitrary, x ( x a < min(1, ɛ/(2 a + 1)) x 2 a 2 < ɛ). Hence δ x ( x a < δ x 2 a 2 < ɛ). Since ɛ > 0 and a are arbitrary, the final result is that a ɛ δ x ( x a < δ x 2 a 2 < ɛ). 2. Consider the proof that f(x) = x 2 is not uniformly continuous. Now the idea is to take x a = δ/2 and use x 2 a 2 = (x+a)(x a) = (2a+δ/2)(δ/2). Here is the proof. With the choice of x a = δ/2 and with a = 1/δ we have that x a < δ and x 2 a 2 1. Hence a x ( x a < δ x 2 a 2 1).

11 1.3. PROOFS FROM ANALYSIS 5 Since δ > 0 is arbitrary, it follows that δ a x ( x a < δ x 2 a 2 1). Finally we conclude that ɛ δ a x ( x a < δ x 2 a 2 ɛ). It is a general fact that f uniformly continuous implies f continuous. This is pure logic; the only problem is to interchange the δ quantifier with the a quantifier. This can be done in one direction. Suppose that δ a A(δ, a). Temporarily suppose that δ is a name for the number that exists, so that aa(δ, a). In particular, A(δ, a ). It follows that δa(δ, a ). This conclusion does not depend on the name, so it follows from the original supposition. Since a is arbitrary, it follows that a δ A(δ, a). What goes wrong with the converse argument? Suppose that a δ A(δ, a). Then δ A(δ, a ). Temporarily suppose A(δ, a ). The trouble is that a is not arbitrary, because something special has been supposed about it. So the generalization is not permitted. Problems 1. A sequence of functions f n converges pointwise (on some set of real numbers) to f as n tends to infinity if x ɛ N n(n N f n (x) f(x) < ɛ). Here the restrictions are that x is in the set and ɛ > 0. Show that for f n (x) = x n and for suitable f(x) there is pointwise convergence on the closed interval [0, 1]. 2. A sequence of functions f n converges uniformly (on some set of real numbers) to f as n tends to infinity if ɛ N x n(n N f n (x) f(x) < ɛ). Show that for f n (x) = x n and the same f(x) the convergence is not uniform on [0, 1]. 3. Show that uniform convergence implies pointwise convergence. 4. Show that if f n converges uniformly to f and if each f n is continuous, then f is continuous. Hint: The first hypothesis is ɛ N x n (n N f n (x) f(x) < ɛ). Deduce that N x n (n N f n (x) f(x) < ɛ /3). Temporarily suppose x n (n N f n (x) f(x) < ɛ /3). The second hypothesis is n a ɛ δ x ( x a < δ f n (x) f n (a) < ɛ). Deduce that δ x ( x a < δ f N (x) f N (a) < ɛ /3). Temporarily suppose that x ( x a < δ f N (x) f N (a) < ɛ /3). Suppose x a < δ. Use the temporary suppositions above to deduce that f(x) f(a) < ɛ. Thus x a < δ f(x) f(a) < ɛ. This is well on the way to the desired conclusion. However be cautious: At this point x is arbitrary, but a is not arbitrary. (Why?) Explain in detail the additional arguments to reach the goal a ɛ δ x( x a < δ f(x) f(a) < ɛ).

12 6 CHAPTER 1. MATHEMATICAL PROOF 1.4 Natural deduction The way of formalizing the rules of logic that corresponds most closely to the practice of mathematical proof is natural deduction. Natural deduction proofs are constructed so that they may be read from the top down. (On the other hand, to construct a natural deduction proof, it is often helpful to work from the top down and the bottom up and try to meet in the middle.) In natural deduction each Suppose introduces a new hypothesis to the set of hypotheses. Each matching Thus removes the hypothesis. Each line is a claim that the formula on this line follows logically from the hypotheses above that have been introduced by a Suppose and not yet eliminated by a matching Thus. Here is an example of a natural deduction proof. Say that one wants to show that if one knows the algebraic fact x (x > 0 (x + 1) > 0), then one is forced by pure logic to accept that y (y > 0 ((y + 1) + 1) > 0). Here is the argument, showing every logical step. The comments on the right are not part of the proof. Suppose x(x > 0 (x + 1) > 0) Suppose z > 0 z > 0 (z + 1) > 0 (specialize the hypothesis) (z + 1) > 0 (from the implication) (z + 1) > 0 ((z + 1) + 1) > 0 (specialize the hypothesis again) ((z + 1) + 1) > 0 (from the implication) Thus z > 0 ((z + 1) + 1) > 0 (introducing the implication) y (y > 0 ((y + 1) + 1) > 0 (generalizing) Notice that the indentation makes the hypotheses in force at each stage quite clear. On the other hand, the proof could also be written in narrative form. It could go like this. Suppose that for all x, if x > 0 then (x + 1) > 0. Suppose z > 0. By specializing the hypothesis, obtain that if z > 0, then (z + 1) > 0. It follows that (z +1) > 0. By specializing the hypothesis again, obtain that if (z +1) > 0, then ((z + 1) + 1) > 0. It follows that ((z + 1) + 1) > 0. Thus if z > 0, then ((z + 1) + 1) > 0. Since z is arbitrary, conclude that for all y, if (y > 0, then ((y + 1) + 1) > 0). Mathematicians usually write in narrative form, but it is useful to practice proofs in outline form, with proper indentation to show the subarguments. The following pages give the rules for natural deduction. In each rule there is a connective or quantifier that is the center of attention. It may be in the hypothesis or in the conclusion. The rule shows how to reduce an argument involving this logical operation to one without the logical operation. (To accomplish this, the rule needs to be used just once, except in two cases involving the substitution of terms. If it were not for these two exceptions, mathematics would be simple indeed.)

13 1.4. NATURAL DEDUCTION 7 Conjunction rules: A B A B and in hypothesis and in hypothesis A B A B and in conclusion Universal rules: x A(x) A(t) all in hypothesis Note: This rule may be used repeatedly with various terms. If z is a variable that does not occur free in a hypothesis in force or in x A, then A(z) x A(x) all in conclusion Note: The restriction on the variable is usually signalled by an expression such as since z is arbitrary, conclude x A(x).

14 8 CHAPTER 1. MATHEMATICAL PROOF Implication rules: A B A B implies in hypothesis Note: This rule by itself is an incomplete guide to practice, since it may not be clear how to prove A. A template that always works is provided below. Suppose A B Thus A B implies in conclusion Negation rules: A A not in hypothesis Note: This rule by itself is an incomplete guide to practice, since it may not be clear how to prove A. A template that always works is provided below. Suppose A Thus A not in conclusion

15 1.4. NATURAL DEDUCTION 9 Disjunction rules: A B Suppose A C Instead suppose B C Thus C or in hypothesis A A B or in conclusion together with B A B or in conclusion Note: This rule by itself is an incomplete guide to practice, since it may not be clear how to prove A or how to prove B. A template that always works is provided below.

16 10 CHAPTER 1. MATHEMATICAL PROOF Existential rules: If z is a variable that does not occur free in a hypothesis in force, in x A, or in C, then x A(x) Suppose A(z) C Thus C exists in hypothesis Note: The restriction on the variable could be signalled by an expression such as since z is arbitrary, conclude C on the basis of the existential hypothesis x A(x). A(t) x A(x) exists in conclusion Note: This rule by itself is an incomplete guide to practice, since it may not be clear how to prove A(t). A template that always works is provided below. This template shows in particular how the rule may be used repeatedly with various terms. Mathematicians tend not to use the exists in hypothesis rule explicitly. They simply suppose that some convenient variable may be used as a name for the thing that exists. They reason with this name up to a point at which they get a conclusion that no longer mentions it. At this point they just forget about their temporary supposition. One could try to formalize this procedure with a rule something like the following. Abbreviated existential rule: If z is a variable that does not occur free in a hypothesis in force, in x A, or in C, then x A(x) Temporarily suppose A(z) C From this point on treat C as a consequence of the existential hypothesis without the temporary supposition or its temporary consequences. In case of doubt, it is safer to use the original rule!

17 1.4. NATURAL DEDUCTION 11 The rules up to this point are those of intuitionistic logic. This is a more flexible form of logic with a very interesting interpretation. Mathematicians find proofs of this type to be natural and direct. However in order to get the full force of classical logic one needs one more rule, the rule of contradiction. The section on natural deduction strategies will demonstrate how this rule may be used in a controlled way. Contradiction rule: Suppose C Thus C by contradiction Note: The double negation law says that A is logically equivalent to A. The rule for negation in conclusion and the double negation law immediately give the contradiction rule.

18 12 CHAPTER 1. MATHEMATICAL PROOF A natural deduction proof is read from top down. However it is often discovered by working simultaneously from the top and the bottom, until a meeting in the middle. The discoverer then obscures the origin of the proof by presenting it from the top down. This is convincing but often not illuminating. Example: Here is a natural deduction proof of the fact that x (x happy x rich) logically implies that x x happy x x rich. Suppose x (x happy x rich) Suppose z happy z rich z happy z rich x x happy x x rich x x happy x x rich Thus x x happy x x rich Here is the same proof in narrative form. Suppose x (x happy x rich). Suppose z happy z rich. Then z happy and hence x x happy. Similarly, z rich and hence x x rich. It follows that x x happy x x rich. Thus (since z is an arbitrary name) it follows that x x happy x x rich on the basis of the original supposition of existence. Example: Here is a natural deduction proof of the fact that x (x happy x rich) logically implies that x x happy x x rich using the abbreviated existential rule. Suppose x (x happy x rich) Temporarily suppose z happy z rich z happy z rich x x happy x x rich x x happy x x rich Here is the same abbreviated proof in narrative form. Suppose x (x happy x rich). Temporarily suppose z happy z rich. Then z happy and hence x x happy. Similarly, z rich and hence x x rich. It follows that x x happy x x rich. Since z is an arbitrary name, this conclusion holds on the basis of the original supposition of existence. Example: Here is a natural deduction proof that y x x y gives x y x y. Suppose y x x y Suppose x x y x y y x y Thus y x y x y x y Here is the same proof in abbreviated narrative form. Suppose y x x y. Temporarily suppose x x y. In particular, x y. Therefore y x y. In fact, since y is arbitrary, this follows on

19 1.5. NATURAL DEDUCTION STRATEGIES 13 the basis of the original existential supposition. Finally, since x is arbitrary, conclude that x y x y. The following problems are to be done using natural deduction. Indent. Justify every logical step. Each step involves precisely one logical operation. The logical operation must correspond to the logical type of the formula. 1. Prove that 2. Suppose Problems x (x rich x happy) ( x x rich x x happy). (1.1) z z 2 0, x y ((x y) 2 0 (2 (x y)) (x 2 + y 2 )). (1.2) Show that it follows logically that x y (2 (x y)) (x 2 + y 2 ). (1.3) 3. Show that the hypotheses n odd n 2 odd, n odd n even, (n 2 odd n 2 even) give the conclusion n 2 even n even. 4. Show that 5. Show that leads logically to x x happy x x happy. (1.4) x y (x likes y x adores y) (1.5) x y x likes y x y x adores y. (1.6) 1.5 Natural deduction strategies A useful strategy for natural deduction is to begin with writing the hypotheses at the top and the conclusion at the bottom. Then work toward the middle. The most important point is to try to use the forall in conclusion rule and the exists in hypothesis rule early in this process of proof construction. This introduces new arbitrary variables. Then one uses the forall in hypothesis rule and the exists in conclusion rule with terms formed from these variables. So it is reasonable to use these latter rules later in the proof construction process. They may need to be used repeatedly. The natural deduction rules as stated above do not have the property that they are reversible. The rules that are problematic are implies in hypothesis, not in hypothesis, or in conclusion, and exists in conclusion. So it is advisable to avoid or postpone the use of these rules. However there are templates that may be used to overcome this difficulty. These have the advantage that they work in all circumstances.

20 14 CHAPTER 1. MATHEMATICAL PROOF A B Suppose C A B implies in hypothesis Thus C by contradiction A Suppose C A Thus C not in hypothesis by contradiction Note: The role of this rule to make use of a negated hypothesis A. When the conclusion C has no useful logical structure, but A does, then the rule effectively switches A for C.

21 1.5. NATURAL DEDUCTION STRATEGIES 15 Suppose (A B) Suppose A A B or in conclusion not in hypothesis Thus A not in conclusion Suppose B A B or in hypothesis not in hypothesis Thus B not in conclusion Thus A B by contradiction Note: A shortcut is to use the DeMorgan s law that says that A B is logically equivalent to ( A B). So if A B leads to a contradiction, then conclude A B. Suppose x A(x) Suppose A(t) x A(x) exists in conclusion not in hypothesis Thus A(t) not in conclusion (may be repeated with various terms) Thus x A(x) by contradiction Note: A shortcut is to use the quantifier DeMorgan s law that says that x A(x) is logically equivalent to ( x A(x)). So if (possibly repeated) use of x A(x) leads to a contradiction, then conclude x A(x).

22 16 CHAPTER 1. MATHEMATICAL PROOF Problems 1. Here is a mathematical argument that shows that there is no largest prime number. Assume that there were a largest prime number. Call it a. Then a is prime, and for every number j with a < j, j is not prime. However, for every number m, there is a number k that divides m and is prime. Hence there is a number k that divides a! + 1 and is prime. Call it b. Now every number k > 1 that divides n! + 1 must satisfy n < k. (Otherwise it would have a remainder of 1.) Hence a < b. But then b is not prime. This is a contradiction. Use natural deduction to prove that logically imply m k (k prime k divides m) (1.7) n k (k divides n! + 1 n < k) (1.8) n (n prime j (n < j j prime)). (1.9) 2. It is a well-known mathematical fact that 2 is irrational. In fact, if it were rational, so that 2 = m/n, then we would have 2n 2 = m 2. Thus m 2 would have an even number of factors of 2, while 2n 2 would have an odd number of factors of two. This would be a contradiction. Use natural deduction to show that i i 2 even-twos (1.10) and give j (j even-twos (2 j) even-twos) (1.11) m n (2 n 2 ) = m 2. (1.12) 1.6 Equality Often equality is thought of as a fundamental logical relation. Manipulations with this concept are very familiar, so there is no need to dwell on it here in detail. However it is worth noting that one could formulate equality rules for natural deduction. Equality rules: For a formula A(z) with free variable z substitution of equals is permitted:

23 1.6. EQUALITY 17 s = t A(s) A(t) equality in hypothesis Everything is equal to itself: t = t equality in conclusion Problems 1. If X is a set, then P (X) is the set of all subsets of X. If X is finite with n elements, then P (X) is finite with 2 n elements. A famous theorem of Cantor states that there is no function f from X to P (X) that is onto P (X). Thus in some sense there are more elements in P (X) than in X. This is obvious when X is finite, but the interesting case is when X is infinite. Here is an outline of a proof. Consider an arbitrary function f from X to P (X). We want to show that there exists a set V such that for each x in X we have f(x) V. Consider the condition that x / f(x). This condition defines a set. That is, there exists a set U such that for all x, x U is equivalent to x / f(x). Call this set S. Let p be arbitrary. Suppose f(p) = S. Suppose p S. Then p / f(p), that is, p / S. This is a contradiction. Thus p / S. Then p f(p), that is, p S. This is a contradiction. Thus f(p) S. Since this is true for arbitrary p, it follows that for each x in X we have f(x) S. Thus there is a set that is not in the range of f. Prove using natural deduction that from U x ((x U x f(x)) ( x f(x) x U)) (1.13) one can conclude that V x f(x) = V. (1.14) 2. Here is an argument that if f and g are continuous functions, then the composite function g f defined by (g f)(x) = g(f(x)) is a continuous function. Assume that f and g are continuous. Consider an arbitrary point a and an arbitrary ɛ > 0. Since g is continuous at f(a ), there exists a δ > 0 such that for all y the condition y f(a ) < δ implies that g(y) g(f(a )) < ɛ. Call it δ 1. Since f is continuous at a, there exists a δ > 0 such that for all x the condition x a < δ implies f(x) f(a ) <

24 18 CHAPTER 1. MATHEMATICAL PROOF δ 1. Call it δ 2. Consider an arbitrary x. Suppose x a < δ 2. Then f(x ) f(a ) < δ 1. Hence g(f(x )) g(f(a )) < ɛ. Thus x a < δ 2 implies g(f(x )) g(f(a )) < ɛ. Since x is arbitrary, this shows that for all x we have the implication x a < δ 2 implies g(f(x)) g(f(a )) < ɛ. It follows that there exists δ > 0 such that all x we have the implication x a < δ implies g(f(x)) g(f(a )) < ɛ. Since ɛ is arbitrary, the composite function g f is continuous at a. Since a is arbitrary, the composite function g f is continuous. In the following proof the restrictions that ɛ > 0 and δ > 0 are implicit. They are understood because this is a convention associated with the use of the variables ɛ and δ. Prove using natural deduction that from and one can conclude that a ɛ δ x ( x a < δ f(x) f(a) < ɛ) (1.15) b ɛ δ y ( y b < δ g(y) g(b) < ɛ) (1.16) a ɛ δ x ( x a < δ g(f(x)) g(f(a)) < ɛ). (1.17) 1.7 Lemmas and theorems In statements of mathematical theorems it is common to have implicit universal quantifiers. For example say that we are dealing with real numbers. Instead of stating the theorem that x y 2xy x 2 + y 2 (1.18) one simply claims that 2uv u 2 + v 2. (1.19) Clearly the second statement is a specialization of the first statement. But it seems to talk about u and v, and it is not clear why this might apply for someone who wants to conclude something about p and q, such as 2pq p 2 + w 2. Why is this permissible? The answer is that the two displayed statements are logically equivalent, provided that there is no hypothesis in force that mentions the variables u or v. Then given the second statement and the fact that the variables in it are arbitrary, the first statement is a valid generalization. Notice that there is no similar principle for existential quantifiers. The statement x x 2 = x (1.20) is a theorem about real numbers, while the statement u 2 = u (1.21)

25 1.8. MORE PROOFS FROM ANALYSIS 19 is a condition that is true for u = 0 or u = 1 and false for all other real numbers. It is certainly not a theorem about real numbers. It might occur in a context where there is a hypothesis that u = 0 or u = 1 in force, but then it would be incorrect to generalize. One cannot be careless about inner quantifiers, even if they are universal. Thus there is a theorem x x < y. (1.22) This could be interpreted as saying that for each arbitrary y there is a number that is smaller than y. Contrast this with the statement x y x < y (1.23) with an inner universal quantifier. This is clearly false for the real number system. The proof rules provided here suffice for every proof in mathematics. This is the famous Gödel completeness theorem. This fact is less useful than one might think, because there is no upper limit to the number of terms that may be used to instantiate a universal hypothesis. Most instantiations are useless, and in complicated circumstances it may be difficult to know the correct one, or even to know that it exists. In practice, the rules are useful only for the construction of small proofs and for verification of a proof after the fact. The way to make progress in mathematics is find concepts that have meaningful interpretations. In order to prove a major theorem, one prepares by proving smaller theorems or lemmas. Each of these may have a rather elementary proof. But the choice of the statements of the lemmas is crucial in making progress. So while the micro structure of mathematical argument is based on the rules of proof, the global structure is a network of lemmas, theorems, and theories based on astute selection of mathematical concepts. 1.8 More proofs from analysis One of the most important concepts of analysis is the concept of open set. This makes sense in the context of the real line, or in the more general case of Euclidean space, or in the even more general setting of a metric space. Here we use notation appropriate to the real line, but little change is required to deal with the other cases. For all subsets V, we say that V is open if a (a V ɛ x ( x a < ɛ x V )). Recall the definition of union of a collection Γ of subsets. This says that for all y we have y Γ if and only if W (W Γ y W ). Here is a proof of the theorem that for all collections of subsets Γ the hypothesis U (U Γ U open) implies the conclusion Γ open. The style of the proof is a relaxed form of natural deduction in which some trivial steps are skipped.

26 20 CHAPTER 1. MATHEMATICAL PROOF Suppose U (U Γ U open). Suppose a Γ. By definition W (W Γ a W ). Temporarily suppose W Γ a W. Since W Γ and W Γ W open, it follows that W open. Since a W it follows from the definition that ɛ x ( x a < ɛ x W ). Temporarily suppose x ( x a < ɛ x W ). Suppose x a < ɛ. Then x W. Since W Γ x W, it follows that W (W Γ x W ). Then from the definition x Γ. Thus x a < ɛ x Γ. Since x is arbitrary, x ( x a < ɛ x Γ). So ɛ x ( x a < ɛ x Γ). Thus a Γ ɛ x ( x a < ɛ x Γ). Since a is arbitrary, a (a Γ ɛ x ( x a < ɛ x Γ)). So by definition Γ open. Thus U (U Γ U open) Γ open. Problems 1. Take the above proof that the union of open sets is open and put it in outline form, with one formula per line. Indent at every Suppose line. Remove the indentation at every Thus line. (However, do not indent at a Temporarily suppose line.) 2. Draw a picture to illustrate the proof in the preceding problem. 3. Prove that for all subsets U, V that (U open V open) U V open. Recall that U V = {U, V } is defined by requiring that for all y that y U V (y U y V ). It may be helpful to use the general fact that for all t, ɛ 1 > 0, ɛ 2 > 0 there is an implication t < min(ɛ 1, ɛ 2 ) (t < ɛ 1 t < ɛ 2 ). Use a similar relaxed natural deduction format. Put in outline form, with one formula per line. 4. Draw a picture to illustrate the proof in the preceding problem. 5. Recall that for all functions f, sets W, and elements t we have t f 1 [W ] f(t) W. Prove that f continuous (with the usual ɛ-δ definition) implies U (U open f 1 [U] open). Use a similar relaxed natural deduction format. 6. It is not hard to prove a lemma that says that {y y b < ɛ} open. Use this lemma and the appropriate definitions to prove that U (U open f 1 [U] open) implies f continuous. Again present this in relaxed natural deduction format.

27 Chapter 2 Sets 2.1 Zermelo axioms Mathematical objects include sets, functions, and numbers. begin with sets. If A is a set, the expression It is natural to t A (2.1) can be read simply t in A. Alternatives are t is a member of A, or t is an element of A, or t belongs to A, or t is in A. The expression t A is often abbreviated t / A and read t not in A. If A and B are sets, the expression is defined in terms of membership by A B (2.2) t (t A t B). (2.3) This can be read simply A subset B. Alternatives are A is included in B or A is a subset of B. (Some people write A B to emphasize that A = B is allowed, but this is a less common convention.) It may be safer to avoid such phrases as t is contained in A or A is contained in B, since here practice is ambiguous. Perhaps the latter is more common. The following axioms are the starting point for Zermelo set theory. They will be supplemented later with the axiom of infinity and the axiom of choice. These axioms are taken by some to be the foundations of mathematics; however they also serve as a review of important constructions. Extensionality A set is defined by its members. For all sets A, B (A B B A) A = B. (2.4) 21

28 22 CHAPTER 2. SETS Empty set Nothing belongs to the empty set. y y /. (2.5) Unordered pair For all objects a, b the unordered pair set {a, b} satisfies y (y {a, b} (y = a y = b)). (2.6) Union If Γ is a set of sets, then its union Γ satisfies x (x Γ A (A Γ x A)) (2.7) Power set If X is a set, the power set P (X) is the set of all subsets of X, so A (A P (X) A X). (2.8) Selection Consider an arbitrary condition p(x) expressed in the language of set theory. If B is a set, then the subset of B consisting of elements that satisfy that condition is a set {x B p(x)} satisfying y (y {x B p(x)} (y B p(y))). (2.9) 2.2 Comments on the axioms Usually in a logical language there is the logical relation symbol = and a number of additional relation symbols and function symbols. The Zermelo axioms could be stated in an austere language in which the only non-logical relation symbol is, and there are no function symbols. The only terms are variables. While this is not at all convenient, it helps to give a more precise formulation of the selection axiom. The following list repeats the axioms in this limited language. However, in practice the other more convenient expressions for forming terms are used. Extensionality A B ( t (t A t B) A = B). (2.10) The axiom of extensionality says that a set is defined by its members. Thus, if A is the set consisting of the digits that occur at least once in my car s license plate 5373, and if B is the set consisting of the odd one digit prime numbers, then A = B is the same three element set. All that matters are that its members are the numbers 7,3,5. Empty set N y y N. (2.11) By the axiom of extensionality there is only one empty set, and in practice it is denoted by the conventional name.

29 2.2. COMMENTS ON THE AXIOMS 23 Unordered pair Union a b E y (y E (y = a y = b)). (2.12) By the axiom of extensionality, for each a, b there is only one unordered pair {a, b}. The unordered pair construction has this name because the order does not matter: {a, b} = {b, a}. Notice that this set can have either one or two elements, depending on whether a = b or a b. In the case when it has only one element, it is written {a} and is called a singleton set. If a, b, c are objects, then there is a set {a, b, c} defined by the condition that for all y y {a, b, c} (y = a y = b y = c). (2.13) This is the corresponding unordered triple construction. The existence of this object is easily seen by noting that both {a, b} and {b, c} exist by the unordered pair construction. Again by the unordered pair construction the set {{a, b}, {b, c}} exists. But then by the union construction the set {{a, b}, {b, c}} exists. A similar construction works for any finite number of objects. Γ U x (x U A (A Γ x A)) (2.14) The standard name for the union is Γ. Notice that = and P (X) = X. A special case of the union construction is A B = {A, B}. This satisfies the property that for all x x A B (x A x B). (2.15) If Γ is a set of sets, then the intersection Γ is defined by requiring that for all x x Γ A (A Γ x A) (2.16) The existence of this intersection follows from the union axiom and the selection axiom: Γ = {x Γ A (A Γ x A)}. There is a peculiarity in the definition of Γ when Γ =. If there is a context where X is a set and Γ P (X), then we can define Γ = {x X A (A Γ x A)}. (2.17) If Γ, then this definition is independent of X and is equivalent to the previous definition. On the other hand, by this definition = X. This might seem strange, since the left hand side does not depend on X. However in most contexts there is a natural choice of X, and this is the definition that is appropriate to such contexts. There is a nice symmetry with the case of union, since for the intersection = X and P (X) =.

30 24 CHAPTER 2. SETS A special case of the intersection construction is A B = {A, B}. This satisfies the property that for all x x A B (x A A B). (2.18) If A X, the complement X \ A is characterized by saying that for all x x X \ A (x X x / A). (2.19) The existence again follows from the selection axiom: X \ A = {x X x / A}. Sometimes the complement of A is denoted A c when the set X is understood. The constructions A B, A B, Γ, Γ, and X \A are means of producing objects that have a special relationship to the corresponding logical operations,,,,. A look at the definitions makes this apparent. Two sets A, B are disjoint if A B =. (In that case it is customary to write the union of A and B as A B.) More generally, a set Γ P (X) of sets is disjoint if for each A in Γ and B Γ with A B we have A B =. A partition of X is a set Γ P (X) such that Γ is disjoint and / Γ and Γ = X. Power set X P A (A P t (t A t X)). (2.20) The power set is the set of all subsets of X, and it is denoted P (X). Since a large set has a huge number of subsets, this axiom has strong consequences for the size of the mathematical universe. Selection The selection axiom is really an infinite family of axioms, one for each formula p(x) expressed in the language of set theory. B S y (y S (y B p(y))). (2.21) The selection axiom says that if there is a set B, then one may select a subset {x B p(x)} defined by a condition expressed in the language of set theory. The language of set theory is the language where the only non-logical relation symbol is. This is why it is important to realize that in principle the other axioms may be expressed in this limited language. The nice feature is that one can characterize the language as the one with just one non-logical relation symbol. However the fact that the separation axiom is stated in this linguistic way is troubling for one who believes that we are talking about a Platonic universe of sets. Of course in practice one uses other ways of producing terms in the language, and this causes no particular difficulty. Often when the set B is understood the set is denoted more simply as {x p(x)}. In the defining condition the quantified variable is implicitly restricted to range over B, so that the defining condition is that for all y we have y {x p(x)} p(y).

31 2.3. ORDERED PAIRS AND CARTESIAN PRODUCT 25 The variables in the set builder construction are bound variables, so, for instance, {u p(u)} is the same set as {t p(t)}. The famous paradox of Bertrand Russell consisted of the discovery that there is no sensible way to define sets by conditions in a completely unrestricted way. Thus if there were a set a = {x x / x}, then a a would be equivalent to a / a, which is a contradiction. Say that it is known that for every x in A there is another corresponding object φ(x) in B. Then another useful notation is This can be defined to be the set {φ(x) B x A p(x)}. (2.22) {y B x (x A p(x) y = φ(x)}. (2.23) So it is a special case. Again, this is often abbreviated as {φ(x) p(x)} when the restrictions on x and φ(x) are clear. In this abbreviated notion one could also write the definition as {y x (p(x) y = φ(x))}. Problems 1. Say X has n elements. How many elements are there in P (X)? 2. Say X has n elements. Denote the number of subsets of X with exactly k elements by ( ( n k). Show that n ) ( 0 = 1 and n n) = 1 and that ) ) ( n 1 ( n k = ( n 1 k 1 Use this to make a table of ( n k) up to n = 7. + k ). (2.24) 3. Say that X has n elements. Denote the number of partitions of X into exactly k non-empty disjoint subsets by S(n, k). This is a Stirling number of the second kind. Show that S(n, 1) = 1 and S(n, n) = 1 and S(n, k) = S(n 1, k 1) + ks(n 1, k). (2.25) Use this to make a table of S(n, k) up to n = Ordered pairs and Cartesian product There is also a very important ordered pair construction. If a, b are objects, then there is an object (a, b). This ordered pair has the following fundamental property: For all a, b, p, q we have (a, b) = (p, q) (a = p b = q). (2.26) If y = (a, b) is an ordered pair, then the first coordinates of y is a and the second coordinate of y is b.

32 26 CHAPTER 2. SETS Some mathematicians like to think of the ordered pair (a, b) as the set (a, b) = {{a}, {a, b}}. The purpose of this rather artificial construction is to make it a mathematical object that is a set, so that one only needs axioms for sets, and not for other kinds of mathematical objects. However this definition does not play much of a role in mathematical practice. There are also ordered triples and so on. The ordered triple (a, b, c) is equal to the ordered triple (p, q, r) precisely when a = p and b = q and c = r. If z = (a, b, c) is an ordered triple, then the coordinates of z are a, b and c. One can construct the ordered triple from ordered pairs by (a, b, c) = ((a, b), c). The ordered n-tuple construction has similar properties. There are degenerate cases. There is an ordered 1-tuple (a). If x = (a), then its only coordinate is a. Furthermore, there is an ordered 0-tuple ( ) = 0 =. Corresponding to these constructions there is a set construction called Cartesian product. If A, B are sets, then A B is the set of all ordered pairs (a, b) with a A and b B. This is a set for the following reason. Let U = A B. Then each of {a} and {a, b} belongs to P (U). Therefore the ordered pair (a, b) belongs to P (P (U)). This is a set, by the power set axiom. So by the selection axiom A B = {(a, b) P (P (U)) a A b B} is a set. One can also construct Cartesian products with more factors. Thus A B C consists of all ordered triples (a, b, c) with a A and b B and c C. The Cartesian product with only one factor is the set whose elements are the (a) with a A. There is a natural correspondence between this somewhat trivial product and the set A itself. The correspondence is that which associates to each (a) the corresponding coordinate a. The Cartesian product with zero factors is a set 1 = {0} with precisely one element 0 =. There is a notion of sum of sets that is dual to the notion of product of sets. This is the disjoint union of two sets. The idea is to attach labels to the elements of A and B. Thus, for example, for each element a of A consider the ordered pair (0, a), while for each element b of B consider the ordered pair (1, b). Then even if there are elements common to A and B, their tagged versions will be distinct. Thus the sets {0} A and {1} B are disjoint. The disjoint union of A and B is the set A + B such that for all y y A + B (y {0} A y {1} B). (2.27) One can also construct disjoint unions with more summands in the obvious way. 2.4 Relations and functions A relation R between sets A and B is a subset of A B. A function (or mapping F from A to B is a relation with the following two properties: x y (x, y) F. (2.28) y y ( x ((x, y) F (x, y ) F ) y = y ). (2.29)